1. ## Surface integral problem

I'm having a little trouble with finishing a surface integral problem. The problem statement is this:

F(x,y,z) = ix + yj + zk, where S is the hemisphere z = (a^2 - x^2 - y^2)^(1/2).

The answer is supposed to be 2pi*a^3, but I'm stuck right after switching to polar coordinates. I've attached a picture of what I've done so far.

Thanks for any help!

2. ## Re: Surface integral problem

Just a quick look: $\vec{T}_s$ is wrong. It should be $\vec{T}_s = (1,0,-s(a^2-s^2-t^2)^{-1/2})$ (and similarly for $\vec{T}_t$).

I took a closer look. It won't affect your calculations. You have:

$\displaystyle \int\int_D\left[(s^2+t^2)(a^2-s^2-t^2)^{-1/2} + (a^2-s^2-t^2)^{1/2}\right]dsdt$

I will rewrite this:

$\dfrac{s^2+t^2}{(a^2-s^2-t^2)^{1/2}} + (a^2-s^2-t^2)^{1/2} = \dfrac{s^2+t^2 + (a^2-s^2-t^2)}{(a^2-s^2-t^2)^{1/2}}$

Simplifying, you get:

$\dfrac{a^2}{(a^2-s^2-t^2)^{1/2}}$

So, your whole integral simplifies to:

$\displaystyle \int\int_D \dfrac{a^2}{(a^2-s^2-t^2)^{1/2}}dsdt$

Does that help?

3. ## Re: Surface integral problem

You are correct! But it's just a typo, the Normal vector should still be the right one.

4. ## Re: Surface integral problem

Originally Posted by gralla55
You are correct! But it's just a typo, the Normal vector should still be the right one.
Very true. I just edited my post above.

5. ## Re: Surface integral problem

And, you don't need to convert to polar coordinates if you don't want. You can use:

$\displaystyle \iint_D \dfrac{a^2}{\sqrt{a^2-s^2-t^2}}dsdt = a^2 \int_{-a}^a \left( \int_{-\sqrt{a^2-t^2}}^{\sqrt{a^2-t^2}} \dfrac{ds}{\sqrt{(\sqrt{a^2-t^2})^2-s^2}} \right)dt$

The inner integral is just $\left. \arcsin\left(\dfrac{s}{\sqrt{a^2-t^2}}\right) \right]_{-\sqrt{a^2-t^2}}^{\sqrt{a^2-t^2}} = \arcsin(1)-\arcsin(-1) = \pi$.

6. ## Re: Surface integral problem

Thanks a lot!! That last one I would never have spotted on my own, I guess I still need lots of practice.